Solucionario De Introduccion A La Termodinamica En Ingenieria Quimica Van Ness 7ma Ed Pdfpdf BEST 📂

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Solucionario De Introduccion A La Termodinamica En Ingenieria Quimica Van Ness 7ma Ed Pdfpdf BEST 📂

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Solucionario De Introduccion A La Termodinamica En Ingenieria Quimica Van Ness 7ma Ed Pdfpdf

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I would use the following:
gsub( “[A-Z]+”, “\\U\\1”, text, perl=TRUE)

Does “like” in differential entropy depend on log of the distribution?

Suppose $X$ follows a distribution $F_X(x)$ and density function $f_X(x)$. $X$ is an integer value, therefore we could have $F_X(x)=0.0001 \space \forall x$.
But now suppose we want to take differential entropy of $X$.
If we use $H[X] = -\int_a^b \log(f_X(x))dx$ (here $a=0, b=4$) and assume $f_X(x)$ is $0.0001$ for all $x$, we get $H[X]=0$. But in differential entropy context, we are supposed to use $H[X]=-\int_a^b f_X(x) \log(f_X(x))dx$
Now, from the first two approaches, entropy of discrete random variable should be $0$. But from the latter approach, discrete random variable should have entropy $\log2$.
So why is this so? Should I use an ordinary integral, and how? What happens when $a=0$ or $b=4$?

$H[X]$ is always a measure of uncertainty with respect to all numbers between $a$ and $b$.

If $a=b=0$ then $H[X]=0$ because the function $f_X(x)$ is zero for all $x$ and the function $-\log(f_X(x))$ is then equal to $+\infty$ (unbounded from above) for all $x$.
If $a\leq x\leq b$ then $f_X(x)\leq f_X(a)\

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